package com.freetymekiyan.algorithms.level.easy;

/**
 * Given a binary tree and a sum, determine if the tree has a root-to-leaf path
 * such that adding up all the values along the path equals the given sum.
 *
 * For example:
 * Given the below binary tree and sum = 22,
 *               5
 *              / \
 *             4   8
 *            /   / \
 *           11  13  4
 *          /  \      \
 *         7    2      1
 *
 * return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
 *
 * Tags: Tree, DFS
 */
class PathSum {

  /**
   * Substract root value from sum every time
   * Return leaf node with sum == 0
   * Or result in left subtree or right subtree
   */
  public boolean hasPathSum(TreeNode root, int sum) {
    if (root == null) return false; // root == null
    sum -= root.val; // update sum
    // leaf? sum == 0? left subtree? right subtree?
    return root.left == null && root.right == null && sum == 0 || hasPathSum(root.left, sum) || hasPathSum(root.right, sum);
  }

  public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int x) {
      val = x;
    }
  }

}